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  • The remainder, when    is divided by 49, is<div class='qna-

The remainder, when  19^{200}+23^{200}  is divided by 49, is

Option: 1

29


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

$$ \begin{aligned} & 19^{200}+23^{200} \quad a^{n+} b^n \\ & 19^3=6859=140 \times 49-1 \\ & =49 \lambda-1 \\ & \left(19^3\right)^{66}=(49 \lambda-1)^{66} \end{aligned}
So, Remainder of 19^{198} divided by 49 is (-1)^{66}=1

19^2=361 gives remainder 18
So, 19^{200} gives remainder 18
23^2 gives remainder 39
(23)^3gives remainder 15
(23)^4 gives remainder 2
\left((23)^4\right)^6 gives remainder (2)^6=64
\&\, 64 gives remainder 15
(23)^{24} \longrightarrow 15$ $(23)^{25} \longrightarrow 2$ $\left((23)^{25}\right)^8 \longrightarrow(2)^8=256 \longrightarrow 11
So, Total remainder =18+11=29

Posted by

Divya Prakash Singh

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